Question

If sum of three numbers is $4$, sum of their squares is $14$, sum of their cubes is $34$ and their products is $-6$, then sum of products of two numbers at a time is

• A. $-1$
• B. $1$
• C. $27$
• D. $2$

Answer 1

The correct option is B $1$

As we know that
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\Rightarrow 34-3(-6)=4(14-\sum ab)$
$\Rightarrow \sum ab=1$